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Let $$f(x)=\sqrt{1+x}$$ Show that if $x \rightarrow 0$ a limit value does exist. Furtheremore, find the limit value and explain the choice of $\delta$ w.r.t $\epsilon$ when the definitions of a limit should be shown.

I need help with $\epsilon - \delta$-proofs and existence. I thought that showing a limit must exist was the whole point of $\epsilon - \delta$-proofs and that it would also specify the value by giving us a $\delta$ thats equal to some expression with $\epsilon$.

Would someone please walk me through how such a problem would be tackled? I.e how would you first show that a limit does indeed exist, formally. Then, find what it is and explain the choice of $\delta$ in the most simple way.

Thanks!

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2 Answers 2

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It's intuitive to see that, if exists, the limit must be $1$. Let's show that. Let $\epsilon\in\mathbb{R}^+$. Then: $$|\sqrt{1 +x}-1|=\left|\dfrac{(\sqrt{1+x}-1)\cdot(\sqrt{1+x}+1)}{\sqrt{1+x}+1}\right|=\left|\dfrac{x}{\sqrt{1+x}+1}\right|$$

Obviously, $1\leq|\sqrt{1+x}+1|=\sqrt{1+x}+1$ for all $x\in\mathbb{R}$, because of $\sqrt{}$ being a non negative function. Then, $$\left|\dfrac{1}{\sqrt{1+x}+1}\right|\leq 1$$

From this two inequalities we can conclude that $|\sqrt{1+x}-1|\leq \epsilon$ whenever $|x|<\epsilon$. So it suffices to choose $\delta=\epsilon$.

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  • $\begingroup$ Hello, I am a bit unsure about the very last step. How do we conclude the $\delta$ from that inequality? $\endgroup$
    – user831952
    Commented Feb 17, 2021 at 13:18
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    $\begingroup$ We have: $$|\sqrt{1+x}-1|=\left|\dfrac{x}{\sqrt{1+x}+1}\right|=|x|\cdot \left|\dfrac{1}{\sqrt{1+x}+1}\right|\leq |x|\cdot 1=|x|$$ because of the inequality we obtained. Then, in order to have $|\sqrt{1+x}-1|<\epsilon$ it suffices to take $|x|<\epsilon=:\delta$, because we would have $|\sqrt{1+x}-1|\leq |x|<\epsilon$. $\endgroup$
    – R.V.N.
    Commented Feb 17, 2021 at 14:08
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Often times, $\epsilon-\delta$ proofs are written to show that a given element is in-fact what the limit is. In order to show that a sequence is convergent without a-priori knowing the limit, one has to use other lines of reasoning. Say, if you are working in the reals, you can perhaps talk about the sequence being Cauchy, or bounded and monotone in a particular direction. In other general spaces, there are more sophisticated arguments that will help you determine if a sequence converges somewhere or not, and some arguments won't necessarily even tell you what the sequence converges to. Even in $\mathbb{R}$ we have methods that tell us $\lim_{n\to \infty} \sum_{k=1}^n \frac{1}{k^3}$ converges (to what is called Aperi's constant), but we have no closed form for the actual number it coverges to, and thus $\epsilon-\delta$ is of little use in this case.

So the best approach to concrete problems such as the one you have (where the function is explicitly given to you) is to hypothesize a certain limit; and then prove using $\epsilon-\delta$ that said hypothesis holds. In this case, from elementary calculus, you are aware that the limit you seek is going to be equal to 1. Now, we use $\epsilon-\delta$ to show that it is indeed 1.

For an actual walk through on how the $\epsilon - \delta$ proof should go, I recommend looking through @R.V.N's answer. My answer would be more or less a repetition of theirs anyway haha

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  • $\begingroup$ So proving that it does exist can be done in the same proof as showing what it is ought to be? $\endgroup$
    – user831952
    Commented Feb 17, 2021 at 12:43
  • $\begingroup$ Yes, it can -- but you generally have to do it before you even begin your $\epsilon-\delta$. So a general proof structure of showing convergence of a sequence may be: 1. See if you can hypothesize a limit, if you can -- great, use $\epsilon-\delta$ to prove that your hypothesis is correct. 2. If you can not hypothesize a limit, use some other argument to ensure convergence. Often times this argument may also tell you what it MIGHT converge to. This will be your hypothesis, and then you will use $\epsilon-\delta$ to prove the hypothesis. $\endgroup$ Commented Feb 18, 2021 at 5:38
  • $\begingroup$ Okay I see, so in this case guessing the limit and then proofing it is sufficient. $\endgroup$
    – user831952
    Commented Feb 18, 2021 at 11:58
  • $\begingroup$ That is correct! $\endgroup$ Commented Feb 18, 2021 at 17:38

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